The problem statement, all variables and given/known data

Consider the following arrangement:

Calculate the work done by tension on 2kg block during its motion on circular track from point $A$ to point $B$.

The attempt at a solution

We know that work done by a force is product of force and displacement.
We know the displacement of point of application as 4m. How to find the work done by the tension as it is not constant it is variable!

2nd attempt(calculus approach)

The 2 kg block moves along the circle, so its speed is Rdθ/dt. It pulls the string, the length of the string between point O and the block can we obtained with simple geometry at any position θ (ignoring the size of the pulley). The total length of the string is unchanged, so the speed of the 1kg block is dL/dt.

2 Answers
2

Work done by tension on BOTH the blocks can be regarded as 0. This can be said by the Virtual Work Method.

The virtual work method:

Consider that block 1(mass $2kg$) displaces by a certain $d\vec{s_1}$. Infinitesimal work done on the block 1 by tension will be given by $$dW_1 = \vec T.d\vec s_1=Tds_1\cos\theta_1$$
Similarly, for block 2 we can say that $$dW_2=\vec T.d\vec s_2=Tds_2\cos\theta_2$$

Using string constraint, we can say that displacement of each block along the string is zero(because the string is inextensible). So we get $$ds_1\cos\theta_1+ds_2\cos\theta_2=0$$
Notice that I have used the same $\theta$ for each block as in tension because the direction along the string is the direction along tension vector.
Net work done by tension thus becomes $$dW_T=T(ds_1\cos\theta_1+ds_2\cos\theta_2)$$
$$\therefore dW_T=0$$ $$W_T=0$$

The solution to the actual problem:

If we apply $W=\Delta K$ on the system of the two blocks from initial position to the final position where block 1 is at the bottom of the cicrular arc,we get $$m_1g\Delta h_1+m_2g\Delta h_2=\frac 12 m_1 v_1^2+\frac 12 m_2 v_2^2$$ I do not include work done by tension on the system because i proved it to be 0.

We now need to find a relation between $v_1$ and $v_2$. We can do this by applying string constraint. $$v_1\cos\theta=v_2$$ where $\cos\theta=\frac 35$(by geometry).
$$\therefore \frac 35 v_1 = v_2$$ Substituting $v_2$ in terms of $v_1$ in the above equation, we can find $v_1$.
Then applying $W=\Delta K$ on block 1 only we get
$$W_T + W_{gravity} = \frac 12 m_1v_1^2$$ Substitute $W_{gravity}$ and $v_1$ in this equation and find $W_T$.

OK, we got off on the wrong foot here. As I mentioned, I was kind of short on time, and therefore I made a lot of shortcuts that, upon closer reading, are not allowed.

So, let's start over. Given the good answer by udiboy above (which was what I was after initially), I'll complement his answer by following the more complicated approach.

The tension will do positive work because of the drop of the 2 kg mass ($W_2$), and negative work because of the rise of the 1 kg mass ($W_1$). Given that the tensile force is always the same on both ends of the string, the total work done by the tension will be zero, because both will simply cancel each other out.

You can get to this result by evaluating the more general expression for work done by a force:

with $\mathbf{F}$ the tension force vector and $d\mathbf{s}$ the displacement vector, and $C_2$ the circular contour that $m_2 = 2\,\text{kg}$ follows, and $C_1$ the linear contour that $m_1 = 1\,\text{kg}$ follows.

These integrals are not so easy to evaluate, since $\mathbf{F}$ changes in both magnitude and direction. Naturally, the circular tract is the most complex piece of this puzzle, so, some definitions first:

It is fairly straightforward to see that

$$
d\mathbf{s}_2 = R \cdot d\theta \cdot \mathbf{e}_\theta
$$

In general, $|\mathbf{F}|$ depends on $\theta$, so remember that $\mathbf{F} = \mathbf{F}(\theta)$. Since we have a dot product in the integrand, and the second vector only has a component in the $\mathbf{e}_\theta$ direction, the contribution of $\mathbf{F}_R$ to the work $W$ will be zero.

For the first integral, there is a complication due to the varying magnitude of $\mathbf{F}$ and the relation between $\theta$ and $d\mathbf{s}_1$. Since we compute $|\mathbf{F}(\theta)|$ inside the second integrand, it is actually easiest to combine both integrals, and compute the work from a single integration over $\theta$.

For this, we need a relation between $d\mathbf{s}_1$ and $d\theta$. Knowing that the total vertical rise of mass $m_1$ equals